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Each field has only an azimuthal component and is independent of $r$ and $\varphi$. The curl in spherical coordinates therefore reduces toand its $\varphi$ component vanishes. The half-angle identitiesshow in both cases that

Apply part (a) with $g(r)=r^{-3}$. Thenaway from the origin, explicitly confirming that each resulting curl has zero divergence.

Put $\theta =\pi /12$. The double-angle trigonometric identities imply ${\cos }^2(2\theta )=1-{\sin }^2(2\theta )$. Since $2\theta =\pi /6$ is in the first quadrant, the positive square root is required. The half-angle identities then give the requested expression in terms of $\sin (\pi /6)$:

Using $\sin (\pi /6)=1/2$ in the preceding half-angle identities givesTo two decimal places the answer is $0.07$.

Rearrange the first integral from part (iv) and use the half-angle identities:Eliminating $\lambda$ givesSquaring removes the sign of the slope; the rising and falling sides of the same profile have opposite tangent angles but obey the same height relation. Heights and angles must remain on the regular branch with $\cos \psi >0$.